Thanks for taking the time to reply Mr Shorty,
So from the initial data I now have:
x y
75 6.81
83 6.95
95 7.08

u v
0 0
8 0.14
20 0.27

** I think for the below extrapolations, I should only be using CORREL, and LINEST, and ANGLE
SLOPE(V1:V3, U1:U3) = 0.013289
INTERCEPT(V1:V3, U1:U3) = 0.12632 - but shouldn't this equal zero? Or is there a different intercept function to use with linest ( , ,FALSE)
CORREL(V1:V3, U1:U3) = 0.99072 - is there a quick way of inferring the at what value the correlation is poor? or do I need trial and error to determine this?
FORECAST(U1, V1:V3, U1:U3) = 0.012632 - but as for intercept, I think I would want this to be 0 - do I need a different function to use with linest ( , ,FALSE)
FORECAST(U2, V1:V3, U1:U3) = 0.118947 -- As mentioned in my original question - Am I right in assuming this is the easiest way to test for "only keep the either the 2nd or third point (whichever is below the line of best fit)"
FORECAST(U3, V1:V3, U1:U3) = 0.278421

LINEST(V1:V3, U1:U3, FALSE) = 0.01405
Instead of FORECAST, is it better to use
u
0 x 0.01405 = estimated v = 0
8 x 0.01405 = estimated v = 0.11241
20 x 0.01405 = estimated v = 0.28103

ANGLE = DEGREES(ATAN((0.28103 - 0)/(20 - 0))) = 0.80505 Are these the correct values to use in this formula? I can't really use the original u and v values, because that doesn't relate to the calculated line of best fit

Regarding the angle - I am looking for essentially the angle within a triangle, were one side is the range of the weeks in the line of best fit, and another side is the range in price of the line of best fit. But shouldn't the 3 angles within a triangle equal 180 degrees, and if one is a right angle, shouldn't the result for the ANGLE, be a lot higher?? Or am I missing a conversion?